Is (tan^2(θ)+1)/(tan^2(θ))=csc^2(θ) ?

The answer to whether (tan^2(θ)+1)/(tan^2(θ))=csc^2(θ) is True

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prove (tan^2(θ)+1)/(tan^2(θ))=csc^2(θ)

Solution

prove

\frac{tan ^{2} ( θ ) + 1}{tan ^{2} ( θ )} = csc^{2} (θ)

Solution

True

Solution steps

\frac{tan ^{2} ( θ ) + 1}{tan ^{2} ( θ )} = csc^{2} (θ)

Manipulating left side

\frac{tan ^{2} ( θ ) + 1}{tan ^{2} ( θ )}

Express with sin, cos

\frac{1 + tan ^{2} ( θ )}{tan ^{2} ( θ )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

Simplify

\frac{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}} : \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

\frac{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

(\frac{sin ( θ )}{cos ( θ )})^{2} = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{\frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}{\frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{( 1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )} ) cos ^{2} ( θ )}{sin ^{2} ( θ )}

Join

1 + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} : \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

1 + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Convert element to fraction:

1 = \frac{1 cos ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{1 \cdot cos ^{2} ( θ )}{cos ^{2} ( θ )} + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{\frac{c o s ^{2} ( θ ) + s i n ^{2} ( θ )}{c o s ^{2} ( θ )} cos ^{2} ( θ )}{sin ^{2} ( θ )}

Multiply

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )} cos^{2} (θ) : cos^{2} (θ) + sin^{2} (θ)

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )} cos^{2} (θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( cos ^{2} ( θ ) + sin ^{2} ( θ ) ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

Cancel the common factor:

cos^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

Rewrite using trig identities

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

= \frac{1}{\frac{1}{c s c ^{2} ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc^{2} (θ)

csc^{2} (θ)

csc^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (tan^2(θ)+1)/(tan^2(θ))=csc^2(θ) ?
The answer to whether (tan^2(θ)+1)/(tan^2(θ))=csc^2(θ) is True